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==A first application of Kuratowski's
The abovementioned Problem 1 (Schmidt), Problem 2 (Dobbertin), and Problem 3 (Goodearl) were solved simultaneously in the negative in 1998.
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It follows from the previously mentioned works of Schmidt, Huhn, Dobbertin, Goodearl, and Handelman that the ℵ<sub>2</sub> bound is optimal in all three negative results above.
As the ℵ<sub>2</sub> bound suggests, infinite combinatorics are involved. The principle used is [[Kuratowski's
The semilattice part of the result above is achieved ''via'' an infinitary semilattice-theoretical statement URP (''Uniform Refinement Property''). If we want to disprove Schmidt's problem, the idea is (1) to prove that any generalized Boolean semilattice satisfies URP (which is easy), (2) that URP is preserved under homomorphic image under a weakly distributive homomorphism (which is also easy), and (3) that there exists a distributive (∨,0)-semilattice of cardinality ℵ<sub>2</sub> that does not satisfy URP (which is difficult, and uses Kuratowski's
Schematically, the construction in the theorem above can be described as follows. For a set Ω, we consider the partially ordered vector space ''E(Ω)'' defined by generators 1 and ''a<sub>i,x</sub>'', for ''i<2'' and ''x'' in Ω, and relations ''a<sub>0,x</sub>+a<sub>1,x</sub>=1'', ''a<sub>0,x</sub> ≥ 0'', and ''a<sub>1,x</sub> ≥ 0'', for any ''x'' in Ω. By using a [[Skolemization]] of the theory of dimension groups, we can embed ''E(Ω)'' functorially into a [[Refinement monoid|dimension vector space]] ''F(Ω)''. The vector space counterexample of the theorem above is ''G=F(Ω)'', for any set Ω with at least ℵ<sub>2</sub> elements.
This counterexample has been modified subsequently by Ploščica and Tůma to a direct semilattice construction. For a (∨,0)-semilattice, the larger semilattice ''R(S)'' is the (∨,0)-semilattice freely generated by new elements ''t(a,b,c)'', for ''a, b, c'' in ''S'' such that ''c ≤ a ∨ b'', subjected to the only relations ''c=t(a,b,c) ∨ t(b,a,c)'' and ''t(a,b,c) ≤ a''. Iterating this construction gives the ''free distributive extension'' <math>D(S)=\bigcup(R^n(S)\mid n<\omega)</math> of ''S''. Now, for a set Ω, let ''L(Ω)'' be the (∨,0)-semilattice defined by generators 1 and ''a<sub>i,x</sub>'', for ''i<2'' and ''x'' in Ω, and relations ''a<sub>0,x</sub> ∨ a<sub>1,x</sub>=1'', for any ''x'' in Ω. Finally, put ''G(Ω)=D(L(Ω))''.
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